Question

1+tan² A\1+cot²A=[1-tanA\1-cotA]²=tan²A

Answer 1

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Answer 2

LHS:

= 1+tan²A / 1+cot²A

= Sec²A/ Cosec²A

= Sin²A/Cos²A

= tan²A

RHS:

=(1-tanA)²/(1-cotA)²

=(1-sinA/cosA)²/(1-cosA/sinA)²

=[(cosA-sinA)/cosA]²

/ [(sinA-cos)/sinA)²

=(cosA-sinA)²×sin²A

/Cos²A. /(sinA-cosA)²

=1×sin²A

/Cos²A×1. ?? (-1)²= 1

=tan²A

Hence proved

Let me know if there are some corrections...Thanks!